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Question 3
A particle P of mass 4 kg is at rest at the point A on a smooth horizontal plane. At time t = 0, two forces, F₁ = (4i - j) N and F₂ = (λi + μj) N, where λ and μ are... show full transcript
Step 1
Answer
To show that ( \lambda - 3\mu + 7 = 0 ), we need to analyze the forces acting on particle P.
The resultant force ( R ) can be computed as follows:
Since the particle moves in the direction of the vector ( (3i + j) ), we can relate the components:
For the i-component: ( 4 + \lambda = k imes 3 ) for some scalar k.
For the j-component: ( -1 + \mu = k imes 1 ) where we say the same scalar ( k ) is used as both forces are in proportion.
From these equations:
Rearranging gives:
We can formulate another expression by eliminating k:
Thus, simplifying yields:
This proves the required equation.
Step 2
Answer
Given ( \lambda = 2 ), we substitute this value back into the earlier derived equations to find ( \mu ):
From the previous step:
Now we can compute the resultant force at ( t = 4 ):
Using ( F_1 ) and ( F_2 ):
Next, we apply Newton's second law:
Now to find the distance traveled in 4 seconds using the equation of motion: Since the particle starts at rest (u = 0), we have:
= \left(\frac{1}{2} \times \frac{3}{2} \times 16, \frac{1}{2} \times \frac{1}{4} \times 16\right) $$ This simplifies to: $$ s = (12, 2) $$ The position vector for point B is from point A to point B: $$ AB = (12i + 2j) $$ Finally, to find the length of AB: $$ |AB| = \sqrt{(12)^2 + (2)^2} = \sqrt{144 + 4} = \sqrt{148} = 2\sqrt{37} \text{ meters} $$Report Improved Results
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